12 Stationary states

12.1 Stationary Wave functions

A simple class of solutions to Schrödinger’s equation may be found by assuming that the dependence on position and time is factorised, \[\psi(x,t) = \phi(x) T(t) \, .\] Substituting into Schrödinger’s equation we find \[i \hbar \frac{1}{T(t)} \frac{\partial T(t)}{\partial t} = - \frac{\hbar^2}{2m} \frac{1}{\phi(x)} \frac{\partial^2\phi(x)}{\partial x^2} + V(x) \, .\] The left hand side is now a function of \(t\) only, while the right hand side is a function of \(x\) only. Since Schrödinger’s equation holds for all \(x\) and \(t\), we conclude that both sides are equal to the same constant, \[\begin{aligned} i \hbar \frac{1}{T(t)} \frac{\partial T(t)}{\partial t} & = E \\ -\frac{\hbar^2}{2m}\frac{1}{\phi(x)} \frac{\partial^2\phi(x)}{\partial x^2} + V(x) & = E \, . \end{aligned}\] where for now \(E\in \mathbb{C}\). Notice we have implicitly assumed the potential \(V(x)\) is independent of time - this will always be the case in this course.

The first equation can be immediately solved, \[T(t) = e^{-iEt/\hbar} \, ,\] up to a normalisation that we can absorb into \(\phi(x)\). The second equation says that \(\phi(x)\) is an eigenfunction of the Hamiltonian \(\hat H\) with eigenvalue \(E\). In particular, since \(\hat H\) is a Hermitian operator, \(E \in \mathbb{R}\).

• Summary: we have found a class of solutions to Schrödinger’s equation \[\psi(x,t) = \phi(x) e^{-iEt/\hbar}\, ,\] where \(\phi(x)\) is an eigenfunction of the Hamiltonian with energy \(E\), \[\label{e:time-indep-schroedinger} \hat H \, \phi(x) = E \phi(x) \, .\]

  The time-independent Schrödinger equation is obtained from the full Schrödinger equation when the wave function is stationary. Solving that equation is equivalent to solving for a particular eigenstate of the Hamiltonian operator.
  This is the time evolution of an initial wave function \(\psi(x,0) = \phi(x)\). Equation \(\eqref{e:time-indep-schroedinger}\) is also often called the “time-independent Schrödinger equation”.

We have encountered a number of examples in the previous lecture. The key property of such solutions is that probabilities of measurements of any observable are independent of time \(t\), since the contribution from the overall phase \(e^{-iEt/\hbar}\) always cancels out. For example,

• Position: The probability density for position is \[P(x,t) = | \psi(x,t)|^2 = | \phi(x) e^{-iEt/\hbar}|^2 = |\phi(x)|^2 \, .\]

• Energy: The probability to measure energy \(E\) is \[P_E = | \langle \phi,\psi\rangle|^2 = |\langle \phi,\phi\rangle e^{-iEt/\hbar}|^2 = |\langle\phi,\phi\rangle|^2 = 1 \, .\]

For this reason, such solutions are known as “stationary wave functions". They are the unique time-evolution of Hamiltonian eigenfunctions.

12.2 Complete Solution

The stationary wave functions are simple but extremely important. Due to the fact that Schrödinger’s equation is linear, we can build all possible solutions of Schrödinger’s equation by taking linear combinations of stationary wave functions!

Suppose we have an initial wave function \(\psi(x,0)\). We now provide a recipe for constructing the solution of Schrödinger’s equation \(\psi(x,t)\) at later times \(t>0\).

1. Step 1: As always, we first construct an orthonormal basis of Hamiltonian eigenfunctions \(\phi_j(x)\) with eigenvalues \(E_j\). We assume here that the spectrum is discrete and non-degenerate. We then have stationary wave functions, \[\psi_j(x,t) = \phi_j(x) e^{-iE_jt/\hbar} \, .\] These wave functions remain orthonormal for all time \(t\) as the dependence on the phase \(e^{-iE_jt/\hbar}\) cancels out, \[\langle \psi_i , \psi_j \rangle = e^{i(E_i-E_j)t/\hbar} \langle\phi_i , \phi_j \rangle = e^{i(E_i-E_j)t/\hbar}\delta_{ij} = \delta_{ij} \, .\]

2. Step 2: We expand the initial wave function \(\psi(x,0)\) as a linear combination of the orthonormal basis of Hamiltonian eigenfunctions, \[\psi(x,0) = \sum_j c_j \phi_j(x) \, .\] The wave function is correctly normalized if \(\langle \psi , \psi \rangle = \sum_j |c_j|^2 = 1\).

3. Step 3: We promote this to the wave function, \[\begin{aligned} \psi(x,t) & = \sum_j c_j \psi_j(x,t) \\ & = \sum_j c_j \phi_j(x) e^{-i E_j t / \hbar } \, . \end{aligned}\] Since Schrödinger’s equation is linear, this is automatically a solution by the principle of superposition. Furthermore, it coincides with the initial wave function \(\psi(x,0)\) at \(t = 0\). Since Schrödinger’s equation is first order in time, we expect the initial wave function to uniquely determine the solution for \(t>0\). We therefore claim that the solution is unique.

How do probabilities and expectation values depend on time?

• Energy. The probabilities of energy measurements are always independent of time because the phases \(e^{-i E_jt/\hbar}\) cancel out \[\begin{aligned} P_j = | \langle \phi_j , \psi\rangle|^2 = | c_j e^{-i E_j t / \hbar}|^2 = |c_j|^2 \, . \end{aligned}\] The expectation value \(\langle H\rangle = \sum_j P_j E_j\) is therefore also independent of \(t\).

• Position and Momentum. The probability distributions and expectation values of position and momentum will typically vary in time.

Let us see this explicitly in some examples.

12.3 Example: Sum of Two Eigenfunctions

Consider a normalized initial wave function \[\psi(x,0) = c_1\phi_1(x) + c_2\phi_2(x)\] with \(|c_1|^2+|c_2|^2=1\). The wave function at later times is \[\begin{aligned} \psi(x,t) & = c_1\psi_1(x,t) + c_2 \psi(x,t) \\[1ex] & = c_1 \phi_1(x) e^{-iE_1t/\hbar} + c_2 \phi_2(x) e^{-iE_2t/\hbar} \, . \end{aligned}\]

The outcomes of energy measurements and their probabilities are independent of \(t\), \[\begin{aligned} E_1 & \quad : \quad P_1 = |c_1|^2 \\ \nonumber E_2 & \quad : \quad P_2 = |c_2|^2 \, . \end{aligned}\] On the other hand, the probability density is \[\begin{aligned} P(x,t) & = |\psi(x,t)|^2 \\[1ex] & = | c_1 \phi_1(x) e^{-iE_1t/\hbar} + c_2 \phi_2(x) e^{-iE_2t/\hbar} |^2 \\[1ex] & = |c_1|^2|\phi_1(x)|^2 + |c_2|^2|\phi_2(x)|^2 \\[1ex] & \qquad + 2 \mathrm{Re}\left(c_1\bar c_2 \phi_1(x)\overline{\phi_2(x)} e^{-i(E_1-E_2)t/\hbar}\right) \, . \end{aligned}\] Due to the final “interference term", the position probability density and position expectation values oscillate in time with frequency \[\omega = (E_2-E_1)/\hbar \, .\]

12.4 Example: Infinite Square Well

Let us consider the infinite square well \(0<x<L\) with initial wave function \[\psi(x,0) = \frac{1}{\sqrt{2}}(\phi_1(x) + \phi_2(x) )\, .\] The unique solution of Schrödinger’s equation is \[\psi(x,t) = \frac{1}{\sqrt{2}} \left(\phi_1(x)e^{-i E_1t/\hbar} +\phi_2(x)e^{-i E_2t/\hbar} \right) \, .\]

The outcomes of energy measurements and their probabilities are independent of \(t\), \[\begin{aligned} E_1 & \quad : \quad P_1 = \frac{1}{2} \\ \nonumber E_2 & \quad : \quad P_2 = \frac{1}{2} \, . \end{aligned}\] On the other hand, the probability density is \[\begin{aligned} P(x,t) & = | \psi(x,t) |^2 \\ \nonumber & = \frac{1}{2} |\phi_1(x)|^2 + \frac{1}{2} |\phi_2(x)|^2 + \frac{1}{2}\left( \phi_1(x) \overline{\phi_2(x)}e^{-i(E_1-E_2)t/\hbar} + \text{c.c.} \right) \\ & = \frac{1}{L} \left( \sin^2\left( \frac{\pi x}{L} \right) + \sin^2\left( \frac{2\pi x}{L} \right) + 2 \sin\left( \frac{\pi x}{L} \right)\sin\left( \frac{2\pi x}{L} \right)\cos(\omega t) \right) \, , \end{aligned}\] which oscillates with frequency \[\omega = \frac{E_2-E_1}{\hbar} = \frac{3\hbar \pi^2}{2mL^2} \, .\]

12.5 Problems

1. Properties of stationary states:

   Suppose \(\{\phi_j(x)\}\) is an orthonormal basis of eigenfunctions of the Hamiltonian \(\hat H\) with eigenvalues \(\{ E_j\}\). Consider the stationary wavefunctions \[\psi_j(x,t) := e^{-E_j t / \hbar} \phi_j(x) \, .\]

   1. Show that \(\langle\psi_i,\psi_j\rangle = \delta_{ij}\) for all \(t\).

   2. Show that \(\psi(x,t)\) satisfy Schödinger’s equation.

   3. Hence explain why \[\psi(x,t) = \sum_{j} c_j \psi_j(x,t)\] is a solution of Schrödinger’s equation for any \(c_j \in \mathbb{C}\).

   4. What equation do \(c_j\) obey if the wavefunction is normalized, \(\langle\psi,\psi\rangle = 1\)?

   5. Show that the probability of measuring energy \(E_j\) is independent of \(t\).

   Solution ▶ See chapter “Energy Revisited”.

2. TUTORIAL 3
   Time evolution on a circle An orthonormal basis of Hamiltonian eigenfunctions for a free particle on a circle of circumference \(L\) is \[\phi_n(x) = \frac{1}{\sqrt{L}} \exp\left(\frac{2\pi i n x}{L}\right) \qquad n \in \mathbb{Z}\] with energy eigenvalues \[E_n = \frac{\hbar^2}{2m}\left( \frac{2\pi n }{L} \right)^2\, .\]

   

   1. The normalised initial wavefunction is \[\psi(x,0) = \sqrt{\frac{2}{L}} \cos \left(\frac{2\pi x}{L}\right) \, .\] Find the wavefunction \(\psi(x,t)\) at \(t>0\) and explain why the solution is a stationary wavefunction.

   2. Suppose the Hamiltonian operator is deformed to \[\hat H = \frac{1}{2m}(\hat p+\alpha)^2\, ,\] where \(0<\alpha < \frac{\pi\hbar}{L}\) is a constant. Find the new wavefunction \(\psi(x,t)\) at \(t>0\) and explain why the solution is no longer stationary.

   Solution ▶

   1. Following the recipe described in the lectures, we first expand the initial wavefunction as a linear combination \[\begin{aligned} \nonumber \psi(x,0) & = \frac{1}{\sqrt{2L}}\left( e^{2\pi i x/L}+ e^{-2\pi i x/L} \right) \\ & = \frac{1}{\sqrt{2}}\left( \phi_1(x) + \phi_{-1}(x) \right) \, . \end{aligned}\] The wavefunction at later times is therefore \[\begin{aligned} \nonumber \psi(x,t) & = \frac{1}{\sqrt{2}}\left( \phi_1(x) e^{-iE_1t/\hbar}+ \phi_{-1}(x) e^{-iE_{-1}t/\hbar}\right) \\ & = \sqrt{\frac{2}{L}} \cos \left(\frac{2\pi x}{L}\right) e^{-i \frac{2\hbar\pi^2}{mL^2} t} \, , \end{aligned}\] where we have used \[E_1 = E_{-1} = \frac{2\hbar^2\pi^2}{mL^2} \, .\] We could also have noticed that since \(E_1 = E_{-1}\), the initial wavefunction \(\psi(x,0)\) is a Hamiltonian eigenfunction. The time evolution is therefore a stationary solution of Schrödinger’s equation.

   2. The energy eigenvalues become \[E_n = \frac{\hbar^2}{2m}\left(\frac{2\pi n}{L} + \frac{\alpha}{\hbar}\right)^2 \, .\] The wavefunction at later times is now \[\begin{aligned} \nonumber \psi(x,t) & = \frac{1}{\sqrt{2}}\left( \phi_1(x) e^{-iE_1t/\hbar} + \phi_{-1}(x) e^{-iE_{-1}t/\hbar} \right) \\ & = \sqrt{\frac{2}{L}} \cos\left(\frac{2\pi}{L}\left(x-\frac{\alpha t}{m} \right)\right) \exp\left[-i \frac{\hbar t}{2m}\left( \left(\frac{2\pi}{L} \right)^2 + \left(\frac{\alpha}{\hbar}\right)^2 \right)\right] \, . \end{aligned}\] As a consistency check, this reduces to the previous result when \(\alpha \to 0\). Since now \(E_1 \neq E_{-1}\) the initial wavefunction \(\psi(x,0)\) is no longer a Hamiltonian eigenfunction and so the wavefunction at later times is not a stationary solution of Schrödinger’s equation.

3. PROBLEMS CLASS 3
   Time dependence in a square well I The wavefunction of an infinite square well \(0<x<L\) at time \(t=0\) is \[\psi(x,0) =C(\phi_1(x) +\phi_2(x))\]

   1. Write down the normalisation factor \(C\).

   2. Write down the wave function \(\psi(x,t)\) at \(t\geq 0\).

   3. What are the possible outcomes of an energy measurement and their probabilities? Why do they not depend on time?

   4. Show that the position expectation value has the form \[\langle x\rangle = \frac{L}{2} - A \cos(\omega t)\] and determine the constants \(A\) and \(\omega\).

   5. Sketch \(\langle x\rangle\) as a function of \(t\).

   6. How would your sketch change if the initial wavefunction were \[\psi(x,0) = C(\phi_1(x) +\phi_3(x)) \, ?\]

   You may use the integral \(\displaystyle \int_0^\pi dy \, y \, \cos\left(n y\right) = \frac{(-1)^n-1}{n^2}\,\) for \(n\in\mathbb{Z}_{>0}\).

   Solution ▶

   1. The normalisation is \(C = 1 / \sqrt{2}\) up to a constant phase.

   2. The wavefunction at time \(t\geq0\) is \[\psi(x,t) = \frac{1}{\sqrt{2}} (\phi_1(x)e^{-iE_1t/\hbar}+\phi_2(x)e^{-iE_2t/\hbar}) \, .\]

   3. The possible outcomes and probabilities are \[\begin{aligned} E_1 & \quad : \quad P_1 = \frac{1}{2} \\ \nonumber E_2 & \quad : \quad P_2 = \frac{1}{2} \, . \end{aligned}\] The probabilities of energy measurements are always independent of time because the phases \(e^{-iE_j t/\hbar}\) cancel out in computing \(P_j = | \langle \phi_j , \psi \rangle|^2\).

   4. Using the inner product notation, the position expectation value is \[\begin{aligned} \nonumber \langle x \rangle & = \langle \psi, \hat x \, \psi \rangle \\ & = \frac{1}{2} \langle \phi_1e^{-iE_1t/\hbar}+\phi_2e^{-iE_2t/\hbar} , \phi_1e^{-iE_1t/\hbar}+\phi_2e^{-iE_2t/\hbar}\rangle \\ & = \langle \phi_1, \hat x \, \phi_1\rangle + \langle \phi_2, \hat x \, \phi_2\rangle + e^{-i(E_2-E_1)t/\hbar} \langle \phi_1, \hat x \, \phi_2\rangle + e^{-i(E_1-E_2)t/\hbar} \langle \phi_2, \hat x \, \phi_1\rangle \\ & = \langle \phi_1, \hat x \, \phi_1\rangle + \langle \phi_2, \hat x \, \phi_2\rangle + e^{-i\omega t} \langle \phi_1, \hat x \, \phi_2\rangle + e^{i\omega t} \langle \phi_2, \hat x \, \phi_1\rangle \end{aligned}\] where \[\omega = (E_1-E_2)/\hbar \, .\] We know that \[\langle \phi_1,\hat x\, \phi_1\rangle = \langle \phi_2,\hat x\, \phi_2\rangle = \frac{L}{2}\] because the Hamiltonian eigenfunctions are either symmetric or anti-symmetric around the centre of the potential well. Using the definite integral given in the question, the cross term is \[\begin{aligned} \nonumber \langle \phi_1, \hat x \, \phi_2\rangle & = \frac{1}{L} \int^L_0 dx \, x \sin\left( \frac{\pi x}{L} \right)\sin\left( \frac{2\pi x}{L} \right) \\ & = \frac{L}{\pi^2} \int^\pi_0 dy\, y \sin\left( y \right)\sin\left( 2y\right) \\ & = \frac{L}{\pi^2} \int^L_0 dy\, y \left(\cos\left( y\right)-\cos\left(3y \right) \right) \\ \nonumber & = - \frac{L}{\pi^2} \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = - \frac{8 L}{9 \pi^2} \, , \end{aligned}\] with the same result for \(\langle \phi_2, \hat x \, \phi_1\rangle\). Putting these results together, we find \[\langle x \rangle = \frac{L}{2} - \frac{16L}{9 \pi^2} \cos(\omega t)\] and can identify \(A = 16L/9\pi^2\).

   5. The sketch should look like the following.

      

   6. If we replace \(\phi_2(x) \to \phi_3(x)\) in the initial wavefunction, the cross term in the computation of \(\langle x\rangle\) vanishes, \[\begin{aligned} \nonumber \langle \phi_1, \hat x \, \phi_3\rangle & = \frac{1}{L} \int^L_0 dx \, x \sin\left( \frac{\pi x}{L} \right)\sin\left( \frac{3\pi x}{L} \right) \\ & = \frac{L}{\pi^2} \int^\pi_0 dy\, y \sin\left( y \right)\sin\left( 3y\right) \\ & = \frac{L}{\pi^2} \int^L_0 dy\, y \left(\cos\left( 2y\right)-\cos\left(4y \right) \right) \\ \nonumber & = 0 \end{aligned}\] and therefore \(\langle x \rangle = L/2\) independent of time.

4. Time dependence in a square well II
   
   The initial wavefunction in an infinite square well \(0<x<L\) is \[\psi(x,0) = C \sin^3\left(\frac{\pi x}{L}\right) \, .\]

   1. Express \(\psi(x,0)\) as a linear combination of an orthonormal basis of Hamiltonian eigenfunctions \(\phi_n(x)\).
      Hint: express \(\sin^3(z)\) as a linear combination of \(\sin(3z)\) and \(\sin(z)\).

   2. Using the orthonormality of \(\phi_n(x)\), determine \(C\).

   3. What are the possible outcomes of an energy measurement and their probabilities?

   4. Hence find the expectation value \(\langle H\rangle\).

   5. Write down the wavefunction \(\psi(x,t)\) for arbitrary later time \(t>0\).

   6. Show that the probability density has the form \[P(x,t) = f(x) + g(x) \cos(\omega t)\] and determine the functions \(f(x)\) and \(g(x)\) and the constant \(\omega\).

   7. Sketch \(P(x,t)\) in the region \(0<x<L\) at times \[t = 0, \frac{\pi}{\omega}, \frac{2\pi}{\omega} \, .\]

   8. Show that \(\langle x \rangle = \displaystyle\frac{L}{2}\) independent of \(t\). Is this consistent with your sketches?

   You may use the integral \(\displaystyle \int_0^\pi dy \, y \, \cos\left(n y\right) = \frac{(-1)^n-1}{n^2}\,\) for \(n\in\mathbb{Z}_{>0}\).

   Solution ▶

   1. We first expand \[\begin{aligned} \sin(x) & = \frac{1}{2i}(e^{ix}-e^{-ix}) \\ \nonumber \sin^3(x) & = -\frac{1}{8i}(e^{ix}-e^{-ix})^3 \\ & = -\frac{1}{8i}(e^{3ix}-3e^{ix} + 3e^{-ix}-e^{-3ix})\\ & = \frac{3}{4}\sin(x) -\frac{1}{4}\sin(3x) \end{aligned}\] and therefore \[\psi(x) = C \sqrt{\frac{L}{2}}\left( \frac{3}{4} \phi_1(x) - \frac{1}{4} \phi_3(x) \right) \, .\]

   2. We now compute inner product using the orthonormal property of the eigenfunctions, \[\langle \psi , \psi \rangle = \frac{L|C|^2}{2}\left(\left(\frac{3}{4}\right)^2 +\left(\frac{1}{4}\right)^2 \right) = \frac{5L}{16} |C|^2 \, .\] For the wavefunction to be normalised, we therefore require \(C = \sqrt{\frac{16}{5L}}\), up to a constant phase.

   3. Substituting the normalization \(C\) back in, the wavefunction is \[\begin{aligned} \psi(x) & = \sqrt{\frac{8}{5}}\left( \frac{3}{4} \phi_1(x) - \frac{1}{4} \phi_3(x) \right) \\ \nonumber & = \sqrt{\frac{9}{10}} \phi_1(x) - \sqrt{\frac{1}{10}} \phi_3(x) \, . \end{aligned}\] The possible outcomes and an energy measurement and their probabilities are therefore given by \[\begin{aligned} E_1 & \quad : \quad P_1 = \frac{9}{10} \\ \nonumber E_3 & \quad : \quad P_3 = \frac{1}{10} \, . \end{aligned}\]

   4. The expectation value of the hamiltonian is \[\langle H \rangle = P_1E_1 + P_2 E_2 = \frac{9}{10} E_1+\frac{1}{10} E_3 = \frac{9\hbar^2\pi^2}{10mL^2} \, .\]

   5. The wave function for \(t>0\) is \[\psi(x,0) = \frac{3}{\sqrt{10}} \phi_1(x)e^{-iE_1t/\hbar} - \frac{1}{\sqrt{10}} \phi_3(x) e^{-iE_3t/\hbar} \, .\]

   6. Noting that the wavefunctions \(\phi_n(x)\) are real, the probability density is \[\begin{aligned} P(x,t) & = | \psi(x,t) |^2 \\ \nonumber & = \frac{9}{10} \phi_1(x)^2 + \frac{1}{10} \phi_3(x)^2 - \frac{3}{10} \phi_1(x) \phi_3(x)\left( e^{-i(E_1-E_3)t/\hbar} + \text{c.c.} \right) \\ & = \frac{9}{10} \phi_1(x)^2 + \frac{1}{10} \phi_3(x)^2 - \frac{6}{10} \phi_1(x) \phi_3(x)\cos(\omega t) \, , \end{aligned}\] where \[\omega = \frac{E_3-E_1}{\hbar} = \frac{4\hbar \pi^2}{mL^2} \, .\] Therefore \[\begin{aligned} \nonumber f(x) & =\frac{9}{10} \phi_1(x)^2 + \frac{1}{10} \phi_3(x)^2 = \frac{9}{5L} \sin^2\left( \frac{\pi x}{L} \right) + \frac{1}{5L} \sin^2\left( \frac{3\pi x}{L} \right) \\ g(x) & = - \frac{6}{10} \phi_1(x) \phi_3(x) = - \frac{6}{5L} \sin\left( \frac{\pi x}{L} \right)\sin\left( \frac{3\pi x}{L} \right) \, . \end{aligned}\]

   7. The sketch looks like the following.

      

   8. To compute the expectation value first note that \[\int^L_0 dx \, x \, \phi_n(x)^2 = \frac{L}{2}\] for any \(n>0\) since \(\phi_n(x)^2\) is symmetric around \(x = L/2\). You may check this explicitly using the hint if you are not convinced. The remaining integral we need is \[\begin{aligned} \nonumber \int^L_0 dx \, x \, \phi_1(x) \phi_3(x) dx & = \frac{2}{L} \int^L_0 x \sin\left( \frac{\pi x}{L} \right)\sin\left( \frac{3\pi x}{L} \right) \\ & = \frac{1}{L} \int^L_0 x \left(\cos\left( \frac{2\pi x}{L} \right)-\cos\left( \frac{4\pi x}{L} \right) \right) \\ & = 0\, , \end{aligned}\] using the hint. We therefore find \[\langle x \rangle = \int^L_0 dx\, x\, f(x) = \frac{9}{10} \cdot \frac{L}{2} + \frac{1}{10} \cdot \frac{L}{2} = \frac{L}{2}\] independent of \(t\). This is consistent with the sketch: \(P(x,t)\) is clearly symmetric around \(x = L / 2\) for the values sketched.